过滤、映射、排序和连接

Question

我觉得可能有更好的解决方案，因为我在以下例程（映射和排序）中重复了代码。

它是具有已读（1 或 null）和未读状态 (0) 的消息的任意列表。我在顶部显示未读消息，而在底部显示已读消息，并应用了一些排序和映射，然后在最后连接两个结果。

var unread = data.filter(function(item){
    return item.Read == 0;
}).map(function(item){
    return {Id: item.Id, First: item.First.toLowerCase(), Last: item.Last.toLowerCase()}
}).sort(function(a, b){
    if (a.Last < b.Last) return -1;
    if (a.Last > b.Last) return 1;
    return 0;
});

var read = data.filter(function(item){
    return item.Read == null || item.Read == 1;
}).map(function(item){ // lowercase (first, last) and sort the list by last
    return {Id: item.Id, First: item.First.toLowerCase(), Last: item.Last.toLowerCase()}
}).sort(function(a, b){
    if (a.Last < b.Last) return -1;
    if (a.Last > b.Last) return 1;
    return 0;
});

var finalData = unread.concat(read);

编辑

var input = [
    {Id: 1, First: "John", Last: "B", Read:0},
    {Id: 1, First: "Jane", Last: "C", Read:0},
    {Id: 1, First: "Doe", Last: "D", Read:1},
    {Id: 1, First: "Alpha", Last: "B", Read:1},
    {Id: 1, First: "Beta", Last: "C", Read:null},
    ];

var output = [
    {Id: 1, First: "Alpha", Last: "B", Read:1},
    {Id: 1, First: "Doe", Last: "D", Read:1},
    {Id: 1, First: "Beta", Last: "C", Read: null},
    {Id: 1, First: "John", Last: "B", Read:0}
    {Id: 1, First: "Jane", Last: "C", Read:0},
];

Answer 1

似乎您只想对多个字段进行排序。要首先按读取状态排序，然后按姓氏，然后按名字（忽略大小写），您可以：

var data = [
  {Id: 1, First: "John", Last: "B", Read:0},
  {Id: 1, First: "Jane", Last: "C", Read:0},
  {Id: 1, First: "Doe", Last: "D", Read:1},
  {Id: 1, First: "Alpha", Last: "B", Read:1}
];

data.sort((a, b) =>
  b.Read !== a.Read
    ? b.Read - a.Read
    : a.Last.toLowerCase().localeCompare(b.Last.toLowerCase())
      ? a.Last.toLowerCase().localeCompare(b.Last.toLowerCase())
      : a.First.toLowerCase().localeCompare(b.First.toLowerCase()));
    
console.log(data);

更新

为了处理Read 字段的null 值被（违反直觉地）认为是真实这一事实，您必须引入几个临时变量（let aRead = a.Read != null a.Read : 1 ) 并比较它们，或将比较重写如下：

var data = [
  {Id: 1, First: "John", Last: "B", Read:0},
  {Id: 1, First: "Jane", Last: "C", Read:0},
  {Id: 1, First: "Doe", Last: "D", Read:1},
  {Id: 1, First: "Alpha", Last: "B", Read:1},
  {Id: 1, First: "Beta", Last: "C", Read:null}
];

data.sort((a, b) =>
  b.Read !== a.Read
    ? (b.Read != null ? b.Read : 1) - (a.Read != null ? a.Read : 1)
    : a.Last.toLowerCase().localeCompare(b.Last.toLowerCase())
      ? a.Last.toLowerCase().localeCompare(b.Last.toLowerCase())
      : a.First.toLowerCase().localeCompare(b.First.toLowerCase()));
    
console.log(data);

Answer 2

你可以先sort，再用reduce累加结果：

var data = [
    {Id: 1, First: "John", Last: "B", Read:0},
    {Id: 1, First: "Jane", Last: "C", Read:0},
    {Id: 1, First: "Doe", Last: "D", Read:1},
    {Id: 1, First: "Alpha", Last: "B", Read:1}
];

var result = data.sort(function(a, b){
    if (a.Last < b.Last) return -1;
    if (a.Last > b.Last) return 1;
    return 0;
}).reduce(function(res, o) {
    var newO = {Id: o.Id, First: o.First.toLowerCase(), Last: o.Last.toLowerCase()}; // the object to be pushed to either the read or unread array (if you are not using the old object o, then just remove the property "Read" from it and use it without creating a new object)
    res[o.Read == 0? "unread": "read"].push(newO);  // push o to either the read or unread array, depending on the property "Read"
    return res;
}, {read: [], unread: []});                         // the initiale value will be an object containing two arrays (one for read and one for unread objects)

console.log(result);

Answer 3

只需将您的 map 和 sort 函数移动到它们自己的命名函数中并重用它们：

// helpers
function processItem(item) {
    return {Id: item.Id, First: item.First.toLowerCase(), Last: item.Last.toLowerCase()}
}

function sortItemByLast(a, b) {
    if (a.Last < b.Last) return -1;
    if (a.Last > b.Last) return 1;
    return 0;
};

// process data
var unread = data.filter(function(item){
    return item.Read == 0;
}).map(processItem).sort(sortItemByLast);

var read = data.filter(function(item){
    return item.Read == null || item.Read == 1;
}).map(processItem).sort(sortItemByLast);


var finalData = unread.concat(read);